How do I write #(5/(1*2))+(5/(2*3))+(5/(3*4))+...+(5/n(n+1))+...#in summation notation, and how can I tell if the series converges?

Question

Cameron Alexander · Accepted Answer

#5\sum_{k=1}^n\frac{1}{k(k+1)}#

The given series:

#(5/{1\cdot 2})+(5/{2\cdot 3})+(5/{3\cdot 4})+\ldots+(5/{n(n+1)})#

#=\sum_{k=1}^n\frac{5}{k(k+1)}#

#=5\sum_{k=1}^n\frac{1}{k(k+1)}#

#=5\sum_{k=1}^n(1/k-\frac{1}{k+1})#

#=5((1-1/2)+(1/2-1/3)+(1/3-1/4)+\ldots+(1/n-1/{n+1}))#

#=5(1-1/2+1/2-1/3+1/3-1/4+\ldots+1/n-1/{n+1})#

#=5(1-1/{n+1})#

#\therefore \lim_{n\to \infty}\sum_{k=1}^n\frac{5}{k(k+1)}#

#=\lim_{n\to \infty}5(1-1/{n+1})#

#=5(1-0)#

#=5#

Hence, the given series is converging

William Abdalla · Answer

Please see some comlementary details below.

I added a few more details The partial fraction decomposition is #1/(k(k+1))=A/k+B/(k+1)# #=(A(k+1)+Bk)/(k(k+1))# Compare the numerators #1=A(k+1)+Bk# Let #k=0#, #=>#, #1=A# Let #k=-1#, #=>#, #1=-B# Therefore, #1/(k(k+1))=1/k-1/(k+1)# Therefore, #sum_(k=1)^n1/(k(k+1))=sum_(k=1)^n1/k-sum_(k=1)^n1/(k+1)# #=1+sum_(k=2)^n1/k-(sum_(k=1)^(n-1)1/(k+1))+1/(n+1)# #=1+sum_(k=2)^n1/k-sum_(k=2)^(n)1/(k)-1/(n+1)# #=1-1/(n+1)# #=n/(n+1)# #=1/(1+1/n)# And #lim_(n->oo)sum_(k=1)^n1/(k(k+1))=lim_(n->oo)1/(1+1/n)# #=1# The series converges and the limit is #=5#

Christopher Allers · Answer

undefined The series can be written in summation notation as:

$$ \sum_{n=1}^{\infty} \frac{5}{n(n+1)} $$

To determine if the series converges, you can use the Ratio Test. Let $a_n = \frac{5}{n(n+1)}$. Then:

$$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{5}{(n+1)(n+2)}}{\frac{5}{n(n+1)}} \right| = \lim_{n \to \infty} \left| \frac{n}{n+2} \right| = 1 $$

Since the limit is equal to 1, the Ratio Test is inconclusive. You can try another convergence test, such as the Integral Test or Comparison Test, to determine the convergence of the series.